Theorem annotanannot 847

Description: A conjunction with a negated conjunction. (Contributed by AV, 8-Mar-2022.) (Proof shortened by Wolf Lammen, 1-Apr-2022.)

Assertion

Ref	Expression
annotanannot	⊢ ((𝜑 ∧ ¬ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ ¬ 𝜓))

Proof of Theorem annotanannot

Step	Hyp	Ref	Expression
1		ibar 537	. . . 4 ⊢ (𝜑 → (𝜓 ↔ (𝜑 ∧ 𝜓)))
2	1	bicomd 226	. . 3 ⊢ (𝜑 → ((𝜑 ∧ 𝜓) ↔ 𝜓))
3	2	notbid 321	. 2 ⊢ (𝜑 → (¬ (𝜑 ∧ 𝜓) ↔ ¬ 𝜓))
4	3	pm5.32i 584	1 ⊢ ((𝜑 ∧ ¬ (𝜑 ∧ 𝜓)) ↔ (𝜑 ∧ ¬ 𝜓))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209   ∧ wa 400

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-an 401

This theorem is used by:  suppcoss  8201  clwwlknclwwlkdif  30341  0nn0m1nnn0  35612